Simplify the following expression and state the condition under which the simplification is valid. $x = \dfrac{-q^2 + 9q - 8}{3q^2 - 3}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ x = \dfrac {-1(q^2 - 9q + 8)} {3(q^2 - 1)} $ $ x = -\dfrac{1}{3} \cdot \dfrac{q^2 - 9q + 8}{q^2 - 1} $ Next factor the numerator and denominator. $ x = - \dfrac{1}{3} \cdot \dfrac{(q - 1)(q - 8)}{(q - 1)(q + 1)}$ Assuming $q \neq 1$ , we can cancel the $q - 1$ $ x = - \dfrac{1}{3} \cdot \dfrac{q - 8}{q + 1}$ Therefore: $ x = \dfrac{ -q + 8 }{ 3(q + 1)}$, $q \neq 1$